May I ask why sin(theta1) / sin(theta2) = 4/3 ? I can’t see an obvious link
It’s from the equation linking refractive index n to angle of incidence θ: n1 sin θ1 = n2 sin θ2. in this case n1 = 1 and n2 = 4/3.
Hi Lucy, thanks very much for your comment.
This is a really interesting point, and I’m really grateful to you for bringing it up.
I had never before come across the formula refractive index = real depth/apparent depth. There is a reason for that: it’s not generally true. If you look at any derivation of the formula (here’s one example http://saburchill.com/physics/chapters3/0004.html), you will see the phrase “If i and r are small angles” (where i and r are the angles of incidence in air and water). But in general the angles of incidence are not small angles — they are only small angles when you are looking straight down at the fish swimming almost directly beneath you. But, when fishing, if you look straight down you will either see a boat or the bank: the only time you will see a fish directly below you is when you are about to fall into the water, which is not a normal or desirable fishing scenario.
I don’t think the solution I have constructed is incorrect, but do I think that you are right that C is a perfectly good answer. If you consider the case when we are looking directly down then θ1 = θ2 = 0, so cosθ1 = cosθ2 = 1, so we do get the answer 1m, which gives us C.
But if you consider all the other cases, where θ1 > θ2, then cosθ1 < cosθ2, we get an answer of more than 1m, which gives us D.
So, given that this is a multiple choice question, we’ve got a bit of a problem! Ah well, I guess you can take comfort in the fact that there won’t be any multiple choice questions in 2015.
For question 21 isn’t the answer actually C: 1m as refractive index = real depth/apparent depth and 1.33×0.75 = 0.9975? The maths in the solution shown is flawed in the initial set up.
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